# Balancing a Cube on its Corner

1. Jun 16, 2005

### amcavoy

Say that you have a box with mass "M." The base is given by vector "v" and height by "w." If you want to balance it on the corner, what should the angle between "v" and the surface be (image attached)?

I tried to write this out, but not sure if I did it correctly.

I figured the torque must be the same (v and w).

$$F_{\mathbf{v}}||\mathbf{v}||=F_{\mathbf{w}}||\mathbf{w}||$$

I figured out that the mass exerted on each is as follows:

$$M_{\mathbf{v}}=M\cos{(\theta)}$$

$$M_{\mathbf{w}}=M\sin{(\theta)}$$

Solving for $$\theta$$, I came up with:

$$\theta=\arctan{(\frac{||\mathbf{w}||}{||\mathbf{v}||})}$$

Is this correct? Does my reasoning make any sense?

Thanks.

2. Jun 17, 2005

### whozum

Well, first off, a cube has 3 dimensions, so if your talking about rectangles thats a different thing. Also, I'm guessing you mean a uniform rectangle. If so, then the diagonal of the rectangle from the pivoting corner to the opposite corner is given by the vector sum of v and w. From here, the torques on either side of this diagonal must be equivalent and in opposite directions. The only way the torques would balance is if the center of mass is above the pivot point. By symmetry, the center of mass will be at the center point of the rectangle, so the center of the rectangle must be directly above the pivot point.

By realizing the diagonal goes through the center, you can see that only when the diagonal is perpendicular to the x axis will this occur, or

$$(\vec{v} + \vec{w}) \cdot \hat{x} = 0$$

3. Jun 17, 2005

### whozum

I realize I didnt answer your question, to find the angle between v and the x axis, draw the vector sum triangle between v and w, name the resultant vector (which is the diagonal) Z.

The angle between Z and X must be 90 degrees, as we showed earlier, but the angle between Z and V is just the arctangent of W/V.

Angle ZX is 90 degrees, and is comprised by angles ZV + VX, where ZV = arctan W/V, so VX is just

$$VX = ZX - ZV = 90 - arctan\left(\frac{|W|}{|V|}\right)$$

Last edited: Jun 17, 2005
4. Jun 17, 2005

### Gale

how i'd do it:

we know two things, we have two angles $$\phi$$ and $$\theta$$, $$\phi$$ being the (small) angle between x and w, and we know these two angles sum together to equal 90 degress. then because of the symmetry and that the center of gravity lies on the diag, we know that there's an equal amount of stuff in the x direction on each side of the diag.... or that $$|v| cos\theta= |w| cos \phi$$ so, we substitute in $$cos\phi= sin\theta$$ and then solve to get $$\frac {|w|}{|v|} tan\theta = 1.$$ and then $$arctan(\frac{|v|}{|w|})=\theta$$

Last edited: Jun 17, 2005
5. Jun 20, 2005